Question: The equation of a circle $C$ is $x^2+y^2-6y = 0$. What is its center $(h, k)$ and its radius $r$ ?
To find the equation in standard form, complete the square. $(x^2) + (y^2-6y) = 0$ $(x^2) + (y^2-6y+9) = 0 + 0 + 9$ $x^2 + (y-3)^{2} = 9 = 3^2$ Thus, $(h, k) = (0, 3)$ and $r = 3$.